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Posted by **winterWX** on Tuesday, March 13, 2007 at 12:52am.

At the instant the flywheel is turning at 75 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 50 cm from the axis of rotation?

For Further Reading

* physics - drwls, Monday, March 12, 2007 at 6:14am

The flywheel will decelerate at a uniform rate, since the frictional torque and moment of inertia can be sonsidered constant. That rate is

alpha = -(250 rev/min)*(2 pi rad/rev)*(60 min/sec)/[(2.1 hr)*(3600 sec/hr)]

-12.5 rad/s^2

Multiply that "alpha" by the radius to get the tangential component of acceleration for a point on the rim of the wheel. It will not matter what the rpm is at the time. RPM will affect the centripetal acceleration, however.

* physics - winterWX, Tuesday, March 13, 2007 at 12:43am

I tried that and got -6233 mm/s^2 or -6.23 m/s^2 but that is the wrong answer. What am I doing wrong? Thanks.

I don't see any error in the method I suggested. 50 cm x -12.5 rad/s^2 is not -6233 mm/s^2, so I don't know how you got that.

I did 500mm x -12.5 rad/s^2 and got -6250 mm/s^2 which are the units asked for but that is the wrong answer. Not sure what to do now.

Thanks for explaining what you did. It seems OK to me now. I'm sorry I cannot be of further assistance with this question.

In the conversion of min to seconds, how did you conclude that 60 min were in one second?

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